Solutions: Section 2 - Whitman College

Solutions: Section 2.6

The material beginning on page 98, Integrating Factors for Exact Equations, will not be on the exam.

1. Problem 1: (2x + 3) + (2y - 2)y = 0 We want fx = M (x, y) = 2x+3 and fy = N (x, y) = 2y -2. We check if this is possible: My = 0 Nx = 0 Now antidifferentiate M with respect to x:

f (x, y) = M (x, y) dx = 2x + 3 dx = x2 + 3x + g(y)

where g is some unknown function of y. Two ways of proceeding (which are equivalent). I'll list both methods for this problem:

? Try to get f from N , and compare:

f (x, y) = N (x, y) dy = 2y - 2 dy = y2 - 2y + g^(x)

where g^ is an unknown function of x. Comparing this to what we had before, we see that:

f (x, y) = x2 + 3x + y2 - 2y so the implicit solution is: x2 + 3x + y2 - 2y = C NOTE: You can always check your answer! ? Another method: Starting from where we left off,

f (x, y) = x2 + 3x + g(y)

we can see what g needs to be in order for fy = N , or: fy = g (y) = 2y - 2 = N

In that case, g(y) = y2 - 2y, and f (x, y) = x2 + 3x + y2 - 2y. The implicit solution is: x2 + 3x + y2 - 2y = C 2. Problem 3: (3x2 - 2xy + 2) dx + (6y2 - x2 + 3) dy = 0 Check to see if the equation is exact:

My = -2x Nx = -2x So yes. Now we'll antidifferentiate M with respect to x:

f (x, y) = M dx = 3x2 - 2xy + 2 dx = x3 - x2y + 2x + g(y)

Check to see if fy is equal to N :

fy = -x2 + g (y) = 6y2 - x2 + 3

so that g (y) = 6y2 + 3. That gives g(y) = 2y3 + 3y. Put this back in to get the full solution, f (x, y) = c:

x3 - x2y + 2x + 2y3 + 3y = C

3.

Problem

4:

(2xy2

+

2y)

+

(2x2y

+

2x)

dy dx

=0

Check for "exactness":

Now set:

My = 4xy + 2 Nx = 4xy + 2 f (x, y) = M dx = 2xy2 + 2y dx = x2y2 + 2xy + g(y)

And check to see that fy = N : fy = 2x2y + 2x + g (y) = 2x2y + 2x

In this case, g (y) = 0, and we don't need to add g(y). The implicit solution:

x2y2 + 2xy = C

4. Problem 13: (2x - y) dx + (2y - x) dy = 0 Check first: My = -1 = Nx, so the DE is exact. Now, f (x, y) = M dx = 2x - y dx = x2 - xy + g(y)

where we check fy to make it equal to N (x, y): fy = -x + g (y) = 2y - x g (y) = 2y g(y) = y2

Our implicit solution is:

x2 - xy + y2 = C

With the initial condition y(1) = 3, we get:

12 - (1)(3) + 32 = C C = 7

The solution to the IVP is:

x2 - xy + y2 = 7

Now we can isolate y and find the restrictions on x (think quadratic formula in y, parentheses added for emphasis):

y2 + (-x) y + x2 - 7 = 0

Now,

x ? x2 - 4(x2 - 7)

x ? 28 - 3x2

y=

=

2

2

Take the positive root since y(1) = 3.

The restriction on x would be that 28 - 3x2 0. Therefore,

28

28

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